Problem: $\overline{AB}$ = $\sqrt{106}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{106}$ $?$ $ \sin( \angle BAC ) = \frac{9\sqrt{106} }{106}, \cos( \angle BAC ) = \frac{5\sqrt{106} }{106}, \tan( \angle BAC ) = \dfrac{9}{5}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{106}} $ $ \overline{BC}=\sqrt{106} \cdot \sin( \angle BAC ) = \sqrt{106} \cdot \frac{9\sqrt{106} }{106} = 9$